Question

If a, $H_1,H_2,.......,H_n,b$ are in H.P., then $\frac{1}{a},\frac{1}{H_1},\frac{1}{H_2},..........,\frac{1}{H_n},\frac{1}{b}$ are in A.P., whose common difference is $\frac{a-b}{(n+1)ab}$.

• A. True
• B. False

Answer 1

The correct option is A True

$\frac{1}{b}=\frac{1}{a}+(n+2-1)d\Rightarrow d=\frac{1}{n+1}[\frac{1}{b}-\frac{1}{a}]$

$=\left[\frac{a-b}{(n+1)ab}\right]$